Linear DP acwing 898 Number triangle

linear DP AcWing 898. Digital triangle

Original link

AcWing 898. Digital triangle

Algorithm tags

Dynamic programming linear DP

Ideas

Code

#include<bits/stdc++.h>
#define int long long
#define rep(i, a, b) for(int i=a;i<b;++i)
#define Rep(i, a, b) for(int i=a;i>b;--i)
using namespace std;
const int N = 505, INF = 0x3f3f3f3f;
int w[N][N], f[N][N];
int n;
inline int read(){
   int s=0,w=1;
   char ch=getchar();
   while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
   while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
   return s*w;
}
void put(int x) {
    if(x<0) putchar('-'),x=-x;
    if(x>=10) put(x/10);
    putchar(x%10^48);
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    n=read();
    rep(i, 1, n+1){
        rep(j, 1, i+1){
            w[i][j]=read();
        }
    }
    rep(i, 1, n+1){
        f[n][i]=w[n][i];
    }
    Rep(i, n-1, -1){
        rep(j, 1, i+1){
            f[i][j]=max(f[i+1][j], f[i+1][j+1])+w[i][j];
        }
    }
    printf("%lld", f[1][1]);
    return 0;
}

Optimize the code

Because the number can be directly stored in f[N][N], When performing state transition operation, it is sufficient to accumulate , So it can be optimized w[N][N]

#include<bits/stdc++.h>
#define int long long
#define rep(i, a, b) for(int i=a;i<b;++i)
#define Rep(i, a, b) for(int i=a;i>b;--i)
using namespace std;
const int N = 505, INF = 0x3f3f3f3f;
int  f[N][N];
int n;
inline int read(){
   int s=0,w=1;
   char ch=getchar();
   while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
   while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
   return s*w;
}
void put(int x) {
    if(x<0) putchar('-'),x=-x;
    if(x>=10) put(x/10);
    putchar(x%10^48);
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    n=read();
    rep(i, 1, n+1){
        rep(j, 1, i+1){
            f[i][j]=read();
        }
    }
    Rep(i, n-1, -1){
        rep(j, 1, i+1){
            f[i][j]+=max(f[i+1][j], f[i+1][j+1]);
        }
    }
    printf("%lld", f[1][1]);
    return 0;
}

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